AIPMT 2015 · Physics

AIPMT 2015 Physics Questions with Solutions

The AIPMT 2015 paper had 45 Physics questions from 25 chapters.

System of Particles and Rotational Motion had the most questions (4), followed by Current Electricity, Mechanical Properties of Fluids and Thermodynamics with 3 each.

10 questions below have the answer and explanation free; the other 35 are in Premium.

Practise AIPMT 2015 Physics free

Physics questions: 45
Chapters covered: 25
Solved free here: 10 of 45
Easy / Medium / Hard: 3 / 24 / 18

Difficulty is NEET MIND's own tag for each question.

Chapter-wise: AIPMT 2015 Physics

How many questions each chapter had in AIPMT 2015. Open a chapter for its questions from every year.

All 45 AIPMT 2015 Physics questions

In paper order. Try each one, then open the answer where it is shown.

Question 1 (AIPMT 2015, Q136)
AtomsMedium
In the spectrum of hydrogen, the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is:
1. Option A: 5/27
2. Option B: 4/9
3. Option C: 9/4
4. Option D: 27/5
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Correct answer: (A) 5/27
Explanation
Using the Rydberg formula: For the longest wavelength in the Lyman series: $\frac{1}{\lambda_L}=R\left(1-\frac{1}{2^2}\right)=\frac{3R}{4}$ Therefore, $\lambda_L=\frac{4}{3R}$ For the longest wavelength in the Balmer series: $\frac{1}{\lambda_B}=R\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=\frac{5R}{36}$ Therefore, $\lambda_B=\frac{36}{5R}$ Hence, $\frac{\lambda_L}{\lambda_B}=\frac{4}{3R}\times\frac{5R}{36}=\frac{5}{27}$
Question 2 (AIPMT 2015, Q137)
Electromagnetic WavesEasy
The energy of the electromagnetic waves is of the order of $15\,\text{keV}$ . To which part of the electromagnetic spectrum does it belong?
1. Option A: $\gamma$ -rays
2. Option B: X-rays
3. Option C: Infra-red rays
4. Option D: Ultraviolet rays
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Question 3 (AIPMT 2015, Q138)
Electromagnetic InductionHard
An electron moves on a straight-line path XY as shown. The loop $abcd$ is placed adjacent to the path of the electron. What will be the direction of current, if any, induced in the loop?
1. Option A: No current induced
2. Option B: abcd
3. Option C: adcb
4. Option D: The current will reverse its direction as the electron goes past the coil
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Correct answer: (D) The current will reverse its direction as the electron goes past the coil
Explanation
As the electron approaches the loop, the magnetic flux linked with the loop changes and an induced current is produced in one direction. After the electron passes the loop and moves away, the magnetic flux change reverses sign. By Lenz's law, the induced current reverses its direction to oppose the new change in magnetic flux. Therefore, the current reverses direction as the electron goes past the coil.
Question 4 (AIPMT 2015, Q139)
Mechanical Properties of FluidsMedium
The cylindrical tube of a spray pump has radius $R$ , one end of which has fine holes, each of radius $r$ . If the speed of the liquid in the tube is $V$ , the speed of ejection of the liquid through the holes is:
1. Option A: $\dfrac{V^2R}{nr}$
2. Option B: $\dfrac{VR^2}{n^2r^2}$
3. Option C: $\dfrac{VR^2}{nr^2}$
4. Option D: $\dfrac{VR^2}{n^3r^2}$
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Question 5 (AIPMT 2015, Q140)
Mechanical Properties of SolidsMedium
The Young's modulus of steel is twice that of brass. Two wires of the same length and same cross-sectional area, one of steel and another of brass, are suspended from the same roof. If the lower ends of the wires are to be at the same level, the weights added to the steel and brass wires must be in the ratio:
1. Option A: 1 : 1
2. Option B: 1 : 2
3. Option C: 2 : 1
4. Option D: 4 : 1
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Correct answer: (C) 2 : 1
Explanation
For equal extensions: $\Delta l=\frac{Fl}{AY}$ Since the lower ends must remain at the same level, $\Delta l_{steel}=\Delta l_{brass}$ $\frac{W_s l}{AY_s}=\frac{W_b l}{AY_b}$ $\frac{W_s}{W_b}=\frac{Y_s}{Y_b}$ Given: $Y_s=2Y_b$ Therefore, $\frac{W_s}{W_b}=2$ Hence the ratio is $2:1$.
Question 6 (AIPMT 2015, Q141)
Current ElectricityHard
A potentiometer wire of length $L$ and resistance $r$ are connected in series with a battery of e.m.f. $E_0$ and a resistance $r_1$ . An unknown e.m.f. $E$ is balanced at a length $\ell$ of the potentiometer wire. The e.m.f. $E$ will be:
1. Option A: $\dfrac{LE_0r}{(r+r_1)\ell}$
2. Option B: $\dfrac{LE_0r}{\ell r_1}$
3. Option C: $\dfrac{E_0r}{(r+r_1)}\cdot\dfrac{\ell}{L}$
4. Option D: $\dfrac{E_0\ell}{L}$
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Question 7 (AIPMT 2015, Q142)
OscillationsMedium
A particle is executing a simple harmonic motion. Its maximum acceleration is $\alpha$ and maximum velocity is $\beta$ . Then, its time period of vibration will be:
1. Option A: $\dfrac{2\pi\beta}{\alpha}$
2. Option B: $\dfrac{\beta^2}{\alpha^2}$
3. Option C: $\dfrac{\alpha}{\beta}$
4. Option D: $\dfrac{\beta^2}{\alpha}$
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Question 8 (AIPMT 2015, Q143)
Motion in a PlaneMedium
If vectors $\vec{A}=\cos\omega t\,\hat{i}+\sin\omega t\,\hat{j}$ $\vec{B}=\cos\left(\frac{\omega t}{2}\right)\hat{i}+\sin\left(\frac{\omega t}{2}\right)\hat{j}$ are functions of time, then the value of $t$ at which they are orthogonal to each other is:
1. Option A: $t=0$
2. Option B: $t=\dfrac{\pi}{4\omega}$
3. Option C: $t=\dfrac{\pi}{2\omega}$
4. Option D: $t=\dfrac{\pi}{\omega}$
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Correct answer: (D) $t=\dfrac{\pi}{\omega}$
Explanation
For orthogonal vectors: $\vec{A}\cdot\vec{B}=0$ Therefore, $\cos(\omega t)\cos\left(\frac{\omega t}{2}\right)+\sin(\omega t)\sin\left(\frac{\omega t}{2}\right)=0$ Using: $\cos x\cos y+\sin x\sin y=\cos(x-y)$ $\cos\left(\omega t-\frac{\omega t}{2}\right)=0$ $\cos\left(\frac{\omega t}{2}\right)=0$ The least positive solution is: $\frac{\omega t}{2}=\frac{\pi}{2}$ Hence, $t=\frac{\pi}{\omega}$ Therefore, option (D) is correct.
Question 9 (AIPMT 2015, Q144)
WavesHard
A source of sound $S$ emitting waves of frequency $100\,\text{Hz}$ and an observer $O$ are located at some distance from each other. The source is moving with a speed of $19.4\,\text{m s}^{-1}$ at an angle of $60^{\circ}$ with the source-observer line as shown in the figure. The observer is at rest. The apparent frequency observed by the observer (velocity of sound in air $330\,\text{ms}^{-1}$ ) is:
1. Option A: 97 Hz
2. Option B: 100 Hz
3. Option C: 103 Hz
4. Option D: 106 Hz
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Correct answer: (C) 103 Hz
Explanation
Only the component of source velocity along the line joining source and observer contributes to the Doppler effect. $v_s'=v_s\cos60^{\circ}=19.4\times\frac{1}{2}=9.7\,\text{m s}^{-1}$ For a moving source and stationary observer: $f'=f\left(\frac{v}{v-v_s'}\right)$ Substituting: $f'=100\left(\frac{330}{330-9.7}\right)$ $f'\approx100\times1.0303$ $f'\approx103\,\text{Hz}$ Hence, option (C) is correct.
Question 10 (AIPMT 2015, Q145)
System of Particles and Rotational MotionMedium
An automobile moves on a road with a speed of $54\,\text{kmh}^{-1}$ . The radius of its wheels is $0.45\,\text{m}$ and the moment of inertia of a wheel about its axis of rotation is $3\,\text{kg m}^2$ . If the vehicle is brought to rest in $15\,\text{s}$ , the magnitude of average torque transmitted by its brakes to a wheel is:
1. Option A: 2.86 kg m² s⁻²
2. Option B: 6.66 kg m² s⁻²
3. Option C: 8.58 kg m² s⁻²
4. Option D: 10.86 kg m² s⁻²
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Question 11 (AIPMT 2015, Q146)
Moving Charges and MagnetismMedium
A rectangular coil of length $0.12\,\text{m}$ and width $0.10\,\text{m}$ having $50$ turns of wire is suspended vertically in a uniform magnetic field of strength $0.2\,\text{Wbm}^{2}$ . The coil carries a current of $2\,\text{A}$ . If the plane of the coil is inclined at an angle of $30^{\circ}$ with the direction of the magnetic field, the torque required to keep the coil in stable equilibrium will be:
1. Option A: $0.12\,\text{N m}$
2. Option B: $0.15\,\text{N m}$
3. Option C: $0.20\,\text{N m}$
4. Option D: $0.24\,\text{N m}$
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Question 12 (AIPMT 2015, Q147)
Electrostatic Potential and CapacitanceHard
A parallel plate air capacitor has capacitance 'C', distance of separation between plates is 'd' and potential difference 'V' is applied between the plates. The force of attraction between the plates of the parallel plate air capacitor is:
1. Option A: $\dfrac{C^2V^2}{2d^2}$
2. Option B: $\dfrac{C^2V^2}{2d}$
3. Option C: $\dfrac{CV^2}{2d}$
4. Option D: $\dfrac{CV^2}{d}$
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Question 13 (AIPMT 2015, Q148)
Kinetic TheoryMedium
Two vessels separately contain two ideal gases A and B at the same temperature, the pressure of A being twice that of B. Under such conditions, the density of A is found to be 1.5 times the density of B. The ratio of molecular weight of A and B is:
1. Option A: 1/2
2. Option B: 2/3
3. Option C: 3/4
4. Option D: 2
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Correct answer: (C) 3/4
Explanation
For an ideal gas, $P=\frac{\rho RT}{M}$ Therefore, $M=\frac{\rho RT}{P}$ Hence, $\frac{M_A}{M_B}=\frac{\rho_A}{\rho_B}\cdot\frac{T_A}{T_B}\cdot\frac{P_B}{P_A}$ Given: $\frac{\rho_A}{\rho_B}=1.5,\qquad T_A=T_B,\qquad P_A=2P_B$ Thus, $\frac{M_A}{M_B}=1.5\times1\times\frac{1}{2}=\frac{3}{4}$ Hence, option (C) is correct.
Question 14 (AIPMT 2015, Q149)
GravitationMedium
A satellite $S$ is moving in an elliptical orbit around the Earth. The mass of the satellite is very small compared to the mass of the Earth. Then,
1. Option A: The acceleration of $S$ is always directed towards the centre of the Earth.
2. Option B: The angular momentum of $S$ about the centre of the Earth changes in direction, but its magnitude remains constant.
3. Option C: The total mechanical energy of $S$ varies periodically with time.
4. Option D: The linear momentum of $S$ remains constant in magnitude.
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Question 15 (AIPMT 2015, Q150)
Semiconductor ElectronicsEasy
In the given figure, a diode $D$ is connected to an external resistance $R=100\,\Omega$ and an e.m.f. of $3.5\,\text{V}$ . If the barrier potential developed across the diode is $0.5\,\text{V}$ , the current in the circuit will be:
1. Option A: 35 mA
2. Option B: 30 mA
3. Option C: 40 mA
4. Option D: 20 mA
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Question 16 (AIPMT 2015, Q151)
GravitationHard
A remote-sensing satellite of Earth revolves in a circular orbit at a height of $0.25\times10^6\,\text{m}$ above the surface of Earth. If Earth's radius is $6.38\times10^6\,\text{m}$ and $g=9.8\,\text{ms}^{-2}$ , then the orbital speed of the satellite is:
1. Option A: $6.67\,\text{km s}^{-1}$
2. Option B: $7.76\,\text{km s}^{-1}$
3. Option C: $8.56\,\text{km s}^{-1}$
4. Option D: $9.13\,\text{km s}^{-1}$
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Question 17 (AIPMT 2015, Q152)
OscillationsHard
The position vector of a particle $\vec{R}$ as a function of time is given by: $\vec{R}=4\sin(2\pi t)\,\hat{i}+4\cos(2\pi t)\,\hat{j}$ where R is in metres, t is in seconds and î and ĵ denote unit vectors along the x and y directions respectively. Which one of the following statements is wrong for the motion of the particle?
1. Option A: Path of the particle is a circle of radius 4 metre
2. Option B: Acceleration vector is along $-\vec{R}$
3. Option C: Magnitude of acceleration vector is $\dfrac{v^2}{R}$ where $v$ is the velocity of the particle
4. Option D: Magnitude of the velocity of particle is 8 metre/second
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Question 18 (AIPMT 2015, Q153)
WavesMedium
A string is stretched between fixed points separated by $75.0\,\text{cm}$ . It is observed to have resonant frequencies of $420\,\text{Hz}$ and $315\,\text{Hz}$ . There are no other resonant frequencies between these two. The lowest resonant frequency for this string is:
1. Option A: 105 Hz
2. Option B: 155 Hz
3. Option C: 205 Hz
4. Option D: 10.5 Hz
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Correct answer: (A) 105 Hz
Explanation
For a string fixed at both ends, resonant frequencies are: $f_n=n\frac{v}{2L}$ Since there are no resonant frequencies between $315\,\text{Hz} $and$ 420\,\text{Hz} $, they are consecutive harmonics. Therefore,$ $f_{n+1}-f_n=420-315=105\,\text{Hz}$ $But$ $f_{n+1}-f_n=\frac{v}{2L}=f_1$ $Hence the fundamental (lowest resonant) frequency is:$ $f_1=105\,\text{Hz}$ $ Therefore, option (A) is correct.
Question 19 (AIPMT 2015, Q154)
System of Particles and Rotational MotionHard
Point masses $m_1$ and $m_2$ are placed at the opposite ends of a rigid rod of length $L$ and negligible mass. The rod is to be set rotating about an axis perpendicular to it. The position of point $P$ on this rod through which the axis should pass so that the work required to set the rod rotating with angular velocity $\omega_0$ is minimum, is given by:
1. Option A: $x=\dfrac{m_2L}{m_1+m_2}$
2. Option B: $x=\dfrac{m_1L}{m_1+m_2}$
3. Option C: $x=\dfrac{m_1}{m_2} L$
4. Option D: $x=\dfrac{m_2}{m_1} L$
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Question 20 (AIPMT 2015, Q155)
Wave OpticsHard
At the first minimum adjacent to the central maximum of a single-slit diffraction pattern, the phase difference between the Huygens wavelet from the edge of the slit and the wavelet from the mid-point of the slit is:
1. Option A: $\dfrac{\pi}{8}$ radian
2. Option B: $\dfrac{\pi}{4}$ radian
3. Option C: $\dfrac{\pi}{2}$ radian
4. Option D: $\pi$ radian
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Question 21 (AIPMT 2015, Q156)
System of Particles and Rotational MotionMedium
A force $\vec{F}=\alpha\hat{i}+3\hat{j}+6\hat{k}$ is acting at a point $\vec{r}=2\hat{i}-6\hat{j}-12\hat{k}$ . The value of $\alpha$ for which angular momentum about the origin is conserved is:
1. Option A: 1
2. Option B: $-1$
3. Option C: 2
4. Option D: 0
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Question 22 (AIPMT 2015, Q157)
Laws of MotionHard
Two particles A and B move with constant velocities $\vec{u}_1$ and $\vec{u}_2$ . At the initial moment their position vectors are $\vec{r}_1$ and $\vec{r}_2$ respectively. The condition for particles A and B to collide is:
1. Option A: $\vec{r}_1-\vec{r}_2=\vec{u}_1-\vec{u}_2$
2. Option B: $\dfrac{\vec{r}_1-\vec{r}_2}{|\vec{r}_1-\vec{r}_2|}=\dfrac{\vec{u}_2-\vec{u}_1}{|\vec{u}_2-\vec{u}_1|}$
3. Option C: $\vec{r}_1\cdot\vec{u}_1=\vec{r}_2\cdot\vec{u}_2$
4. Option D: $\vec{r}_1\times\vec{u}_1=\vec{r}_2\times\vec{u}_2$
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Question 23 (AIPMT 2015, Q158)
NucleiMedium
A nucleus of uranium decays at rest into nuclei of thorium and helium. Then:
1. Option A: The helium nucleus has less kinetic energy than the thorium nucleus.
2. Option B: The helium nucleus has more kinetic energy than the thorium nucleus.
3. Option C: The helium nucleus has less momentum than the thorium nucleus.
4. Option D: The helium nucleus has more momentum than the thorium nucleus.
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Question 24 (AIPMT 2015, Q159)
Current ElectricityMedium
Two metal wires of identical dimensions are connected in series. If $\sigma_1$ and $\sigma_2$ are the conductivities of the metal wires respectively, the effective conductivity of the combination is:
1. Option A: $\dfrac{\sigma_1\sigma_2}{\sigma_1+\sigma_2}$
2. Option B: $\dfrac{2\sigma_1\sigma_2}{\sigma_1+\sigma_2}$
3. Option C: $\dfrac{\sigma_1+\sigma_2}{2\sigma_1\sigma_2}$
4. Option D: $\dfrac{\sigma_1+\sigma_2}{\sigma_1\sigma_2}$
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Question 25 (AIPMT 2015, Q160)
Dual Nature of Radiation and MatterHard
Light of wavelength $500\,\text{nm}$ is incident on a metal with work function $2.28\,\text{eV}$ . The de Broglie wavelength of the emitted electron is:
1. Option A: ≤ 2.8 × 10⁻¹² m
2. Option B: < 2.8 × 10⁻¹⁰ m
3. Option C: < 2.8 × 10⁻⁹ m
4. Option D: ≥ 2.8 × 10⁻⁹ m
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Question 26 (AIPMT 2015, Q161)
ThermodynamicsHard
$4.0\,\text{g}$ sample of a gas occupies $22.4\,\text{L}$ at NTP. The specific heat capacity of the gas at constant volume is $5.0\,\text{JK}^{-1}\text{mol}^{-1}$ . If the speed of sound in this gas at NTP is $952\,\text{ms}^{-1}$ , then the heat capacity at constant pressure is: (Take gas constant R = 8.3 $\text{JK}^{-1}\text{mol}^{-1}$ )
1. Option A: 8.5 JK⁻¹ mol⁻¹
2. Option B: 8.0 JK⁻¹ mol⁻¹
3. Option C: 7.5 JK⁻¹ mol⁻¹
4. Option D: 7.0 JK⁻¹ mol⁻¹
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Question 27 (AIPMT 2015, Q162)
Alternating CurrentMedium
A series R-C circuit is connected to an alternating voltage source. Consider two situations: (a) When the capacitor is air filled. (b) When the capacitor is mica filled. Current through the resistor is $i$ and voltage across the capacitor is $V$ . Then:
1. Option A: $V_a = V_b$
2. Option B: $V_a < V_b$
3. Option C: $V_a > V_b$
4. Option D: $i_a > i_b$
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Question 28 (AIPMT 2015, Q163)
Laws of MotionHard
A plank with a box on it at one end is gradually raised about the other end. As the angle of inclination with the horizontal reaches $30^\circ$ , the box starts to slip and slides $4.0\,\text{m}$ down the plank in $4.0\,\text{s}$ . The coefficients of static and kinetic friction between the box and the plank will be, respectively:
1. Option A: 0.4 and 0.3
2. Option B: 0.6 and 0.6
3. Option C: 0.6 and 0.5
4. Option D: 0.5 and 0.6
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Question 29 (AIPMT 2015, Q164)
System of Particles and Rotational MotionEasy
Two stones of masses $m$ and $2m$ are whirled in horizontal circles, the heavier one in a radius $\dfrac{r}{2}$ and the lighter one in radius $r$ . The tangential speed of the lighter stone is $n$ times that of the heavier stone when they experience the same centripetal force. The value of $n$ is:
1. Option A: 1
2. Option B: 2
3. Option C: 3
4. Option D: 4
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Question 30 (AIPMT 2015, Q165)
ThermodynamicsMedium
The coefficient of performance of a refrigerator is 5. If the temperature inside the freezer is $-20^\circ\text{C}$ , the temperature of the surroundings to which it rejects heat is:
1. Option A: $21^\circ\text{C}$
2. Option B: $31^\circ\text{C}$
3. Option C: $41^\circ\text{C}$
4. Option D: $11^\circ\text{C}$
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Question 31 (AIPMT 2015, Q166)
ThermodynamicsMedium
An ideal gas is compressed to half its initial volume by means of several processes. Which of the process results in the maximum work done on the gas?
1. Option A: Isothermal
2. Option B: Adiabatic
3. Option C: Isobaric
4. Option D: Isochoric
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Question 32 (AIPMT 2015, Q167)
Work, Energy and PowerMedium
A ball is thrown vertically downwards from a height of $20\,\text{m}$ with an initial velocity $v_0$ . It collides with the ground, loses $50\%$ of its energy in collision and rebounds to the same height. The initial velocity $v_0$ is: (Take $g=10\,\text{ms}^{-2}$ )
1. Option A: $10\,\text{ms}^{-1}$
2. Option B: $14\,\text{ms}^{-1}$
3. Option C: $20\,\text{ms}^{-1}$
4. Option D: $28\,\text{ms}^{-1}$
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Correct answer: (C) $20\,\text{ms}^{-1}$
Explanation
To rebound to a height of $20\,\text{m}$ , the speed immediately after collision must be: $v=\sqrt{2gh}=\sqrt{2\times10\times20}=20\,\text{m s}^{-1}$ Energy just after rebound: $E=\frac{1}{2}mv^2=\frac{1}{2}m(20)^2=200m$ Since $50\% $of the energy is lost during collision, the energy just before collision was:$ $E_{before}=400m$ $Applying energy conservation during the fall:$ $\frac{1}{2}mv_0^2+mgh=400m$ $\frac{1}{2}v_0^2+10\times20=400$ $\frac{1}{2}v_0^2=200$ $v_0=20\,\text{m s}^{-1}$ $ Hence, option (C) is correct.
Question 33 (AIPMT 2015, Q168)
Work, Energy and PowerHard
On a frictionless surface, a block of mass $M$ moving at speed $v$ collides elastically with another block of the same mass $M$ which is initially at rest. After collision the first block moves at an angle $\theta$ to its initial direction and has a speed $\dfrac{v}{3}$ . The second block's speed after the collision is:
1. Option A: $\dfrac{\sqrt{3}}{2}v$
2. Option B: $\dfrac{2\sqrt{2}}{3}v$
3. Option C: $\dfrac{3}{4}v$
4. Option D: $\dfrac{3}{\sqrt{2}}v$
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Correct answer: (B) $\dfrac{2\sqrt{2}}{3}v$
Explanation
Since the collision is elastic, kinetic energy is conserved. Before collision: $K_i=\frac{1}{2}Mv^2$ After collision: $K_f=\frac{1}{2}M\left(\frac{v}{3}\right)^2+\frac{1}{2}MV'^2$ Equating: $v^2=\frac{v^2}{9}+V'^2$ $V'^2=\frac{8}{9}v^2$ $V'=\frac{2\sqrt{2}}{3}v$ Hence, option (B) is correct.
Question 34 (AIPMT 2015, Q169)
Electrostatic Potential and CapacitanceMedium
If potential (in volts) in a region is expressed as $V(x,y,z)=6xy-y+2yz,$ the electric field (in N/C) at the point (1, 1, 0) is:
1. Option A: $-(6\hat{i}+9\hat{j}+\hat{k})$
2. Option B: $-(3\hat{i}+5\hat{j}+3\hat{k})$
3. Option C: $-(6\hat{i}+5\hat{j}+2\hat{k})$
4. Option D: $-(2\hat{i}+3\hat{j}+\hat{k})$
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Question 35 (AIPMT 2015, Q170)
Wave OpticsHard
Two slits in Young's experiment have widths in the ratio $1:25$ . The ratio of intensity at the maxima and minima in the interference pattern, $\dfrac{I_{\max}}{I_{\min}}$ , is:
1. Option A: $\dfrac{4}{9}$
2. Option B: $\dfrac{9}{4}$
3. Option C: $\dfrac{121}{49}$
4. Option D: $\dfrac{49}{121}$
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Question 36 (AIPMT 2015, Q171)
Mechanical Properties of FluidsMedium
The heart of a man pumps $5$ litres of blood through the arteries per minute at a pressure of $150\,\text{mm}$ of mercury. If the density of mercury is $13.6\times10^3\,\text{kg/m}^{3}$ and $g=10\,\text{m/s}^{2}$ , then the power of heart in watt is:
1. Option A: 1.50
2. Option B: 1.70
3. Option C: 2.35
4. Option D: 3.0
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Question 37 (AIPMT 2015, Q172)
Moving Charges and MagnetismHard
A proton and an alpha particle both enter a region of uniform magnetic field $B$ , moving at right angles to the field. If the radii of circular orbits for both the particles are equal and the kinetic energy acquired by the proton is $1\,\text{MeV}$ , then the energy acquired by the alpha particle will be:
1. Option A: 1 MeV
2. Option B: 4 MeV
3. Option C: 0.5 MeV
4. Option D: 1.5 MeV
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Question 38 (AIPMT 2015, Q173)
Semiconductor ElectronicsMedium
The input signal given to a CE amplifier having a voltage gain of 150 is $V_i = 2\cos\left(15t+\frac{\pi}{3}\right).$ The corresponding output signal will be:
1. Option A: $300\cos\left(15t+\frac{4\pi}{3}\right)$
2. Option B: $300\cos\left(15t+\frac{\pi}{3}\right)$
3. Option C: $75\cos\left(15t+\frac{2\pi}{3}\right)$
4. Option D: $2\cos\left(15t+\frac{5\pi}{6}\right)$
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Question 39 (AIPMT 2015, Q174)
Units and MeasurementsHard
In dimension of critical velocity vc of liquid flowing through a tube are expressed as (ηˣ ρʸ rᶻ), where η, ρ and r are the coefficients of viscosity of liquid, density of liquid and radius of the tube respectively, then the values of x, y and z are given by:
1. Option A: 1, 1, 1
2. Option B: 1, −1, −1
3. Option C: −1, −1, 1
4. Option D: −1, −1, −1
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Question 40 (AIPMT 2015, Q175)
Current ElectricityMedium
A circuit contains an ammeter, a battery of 30 V and a resistance 40.8 ohm all connected in series. If the ammeter has a coil of resistance 480 ohm and a shunt of 20 ohm, the reading in the ammeter will be :-
1. Option A: 1 A
2. Option B: 0.5 A
3. Option C: 0.25 A
4. Option D: 2 A
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Question 41 (AIPMT 2015, Q176)
Mechanical Properties of FluidsMedium
Water rises to height $h$ in a capillary tube. If the length of capillary tube above the surface of water is made less than $h$ , then:
1. Option A: Water does not rise at all.
2. Option B: Water rises up to the tip of the capillary tube and then starts overflowing like a fountain.
3. Option C: Water rises up to the top of the capillary tube and stays there without overflowing.
4. Option D: Water rises up to a point a little below the top and stays there.
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Question 42 (AIPMT 2015, Q177)
Ray Optics and Optical InstrumentsHard
In an astronomical telescope in normal adjustment, a straight black line of length $L$ is drawn on the inside part of the objective lens. The eyepiece forms a real image of this line. The length of this image is $l$ . The magnification of the telescope is:
1. Option A: $\dfrac{L}{l}$
2. Option B: $\dfrac{L}{l}+1$
3. Option C: $\dfrac{L}{l}-1$
4. Option D: $\dfrac{L+l}{L-l}$
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Question 43 (AIPMT 2015, Q178)
Thermal Properties of MatterMedium
The value of coefficient of volume expansion of glycerin is $5\times10^{-4}\,\text{K}^{-1}$ . The fractional change in the density of glycerin for a rise of $40^{\circ}\text{C}$ in its temperature is:
1. Option A: 0.010
2. Option B: 0.015
3. Option C: 0.020
4. Option D: 0.025
Show answer & explanationHide answer & explanation
Correct answer: (C) 0.020
Explanation
For a liquid, $V_f=V_i(1+\gamma\Delta T)$ Since density is inversely proportional to volume, $\rho_f=\frac{\rho_i}{1+\gamma\Delta T}$ Fractional decrease in density: $\frac{\rho_i-\rho_f}{\rho_i}=1-\frac{1}{1+\gamma\Delta T}$ For small expansion, $\frac{1}{1+x}\approx 1-x$ Therefore, $\frac{\Delta\rho}{\rho}\approx \gamma\Delta T$ $=(5\times10^{-4})(40)$ $=2.0\times10^{-2}=0.020$ Hence option (C) is correct.
Question 44 (AIPMT 2015, Q179)
Dual Nature of Radiation and MatterMedium
A photoelectric surface is illuminated successively by monochromatic light of wavelength $\lambda$ and $\dfrac{\lambda}{2}$ . If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function of the surface material is: ( $h$ = Planck's constant, $c$ = speed of light)
1. Option A: hc/3λ
2. Option B: hc/2λ
3. Option C: hc/λ
4. Option D: 2hc/λ
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Question 45 (AIPMT 2015, Q180)
Ray Optics and Optical InstrumentsHard
A beam of light consisting of red, green and blue colours is incident on a right-angled prism. The refractive indices of the material of the prism for the red, green and blue wavelengths are 1.39, 1.44 and 1.47 respectively. The prism will:
1. Option A: Separate the red colour part from the green and blue colours.
2. Option B: Separate the blue colour part from the red and green colours.
3. Option C: Separate all the three colours from one another.
4. Option D: Not separate the three colours at all.
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